Question

A charge$\mathit{q}$ is released from rest at the origin, in the presence of a uniform electric field$\mathit{E}\mathbf{=}{\mathbf{E}}_{\mathbf{0}}\widehat{\mathbf{z}}$ and a uniform magnetic field$\mathit{B}\mathbf{=}{\mathbf{B}}_{\mathbf{0}}\widehat{\mathbf{x}}$ . Determine the trajectory of the particle by transforming to a system in Which,$\mathit{E}\mathbf{=}\mathbf{0}$ , finding the path in that system and then transforming back to the original system. Assume${\mathit{E}}_{\mathbf{0}}\mathbf{<}\mathit{c}{\mathit{B}}_{\mathbf{0}}$ .Compare your result with Ex. 5.2.

Short answer

The trajectory of the charge is ‘an elliptical cycloid’ represented by the equation ${\gamma }^2{(y-\upsilon t)}^2+{(z-R)}^2=R^2$.

Step-by-step solution

Given information

The magnitude of the charge released from rest at the origin is$q$ .

The magnitude of the uniform electric field is,$E=E_0\widehat{z}$ .

The magnitude of the uniform magnetic field is, $B=B_0\widehat{x}$.

Momentum of the charged particle

The trajectory of a charged particle in a uniform magnetic field is always a ‘cyclotron motion’.

The formula for the momentum$\left(p\right)$ of the charge$q$ traveling in a cyclotron motion with radius$R$ in a uniform magnetic field$B$ is given by,

$p=qBR$

Radius of the path of the charge 

The initial position of the charge at rest located at the origin can be represented by following diagram,

Here,$\upsilon$represents the velocity of the charge moving away from the origin.

The complete set of transformation rules for the electric field in respective axes,

$\begin{array}{l}{\overline{E}}_x=E_x\\ {\overline{E}}_y=\gamma (E_y-\upsilon B_z)\\ {\overline{E}}_z=\gamma (E_z+\upsilon B_y)\end{array}$

And, the magnetic field in respective axes,

$\begin{array}{l}{\overline{B}}_x=B_x\\ {\overline{B}}_y=\gamma \left(B_y+\frac{\upsilon }{c^2}{E}_z\right)\\ {\overline{B}}_z=\gamma \left(B_y-\frac{\upsilon }{c^2}{E}_y\right)\end{array}$

Rewriting the complete set of transformation rules by putting$x\to y,y\to z,z\to x$,

The electric field is,

$\begin{array}{l}{\overline{E}}_y=E_y\\ {\overline{E}}_z=\gamma (E_z-\upsilon B_x)\\ {\overline{E}}_x=\gamma (E_x+\upsilon B_z)\end{array}$

And, magnetic field,

$\begin{array}{l}{\overline{B}}_y=B_y\\ {\overline{B}}_z=\gamma \left(B_z+\frac{\upsilon }{c^2}{E}_x\right)\\ {\overline{B}}_x=\gamma \left(B_x-\frac{\upsilon }{c^2}{E}_z\right)\end{array}$

This equation gives the electric and magnetic fields in system $\overline{S}$moving in the y direction at speed $\nu$.

The initial value of electric field in co-ordinate form is given by,

$E=(0,0,E_0)$

And, the initial magnetic field in co-ordinate form is given by,

$B=(B_0,0,0)$

So, the initial electric field can be written as,

$\begin{array}{c}{\overline{E}}_y=0\\ {\overline{E}}_z=\gamma (E_0-\upsilon B_0)\\ {\overline{E}}_x=0\end{array}$

Now if the value of electric field is$\overline{E}=0$, then the value of speed$\upsilon$can be calculated as,

$\begin{array}{c}{E}_0-\upsilon B_0=0\\ \upsilon =\frac{E_0}{B_0}\end{array}$

(The condition $\frac{E_0}{B_0}<0$guarantees that there is no problem getting to such a system.)

According to the condition, the value of magnetic field will be,

$\begin{array}{c}{\overline{B}}_y=0\\ {\overline{B}}_z=0\\ {\overline{B}}_x=\gamma \left(B_0-\frac{\upsilon }{c^2}{E}_0\right)\end{array}$

Putting $E_0=\upsilon B_0$and solving,

$\begin{array}{c}{\overline{B}}_x=\gamma B_0\left(1-\frac{{\upsilon }^2}{c^2}\right)\\ {\overline{B}}_x=\gamma B_0\frac{1}{{\gamma }^2}\\ {\overline{B}}_x=\frac{1}{\gamma }{B}_0\end{array}$

In standard form,

$\overline{B}=\frac{1}{\gamma }{B}_0\widehat{x}$

The trajectory of the charge in the relative axis after a certain time$\left(t\right)$rotating with angular velocity$\left(\omega \right)$is represented as,

The trajectory in$\overline{S}$ .Since the particle started out at rest at the origin in $S$, it started out with velocity$-\upsilon \widehat{y}$ in $\overline{S}$. According to the formula for the momentum of the charge, moving in a circle of radius$R$ is given by

$\begin{array}{c}p=qBR\\ \gamma m\upsilon =q\left(\frac{1}{\gamma }{B}_0\right)R\\ R=\frac{m{\gamma }^2\upsilon }{q{B}_0}\end{array}$

Here,$m$ is the mass of the charge.

Trajectory of the charge

Then using the transformation rules, the actual trajectory of the particle is given by,

$\begin{array}{c}\overline{x}=0\\ \overline{y}=-R\sin \omega \overline{t}\\ \overline{z}=R(1-\cos \omega \overline{t})\end{array}$

Here, the angular velocity, $\omega =\frac{\upsilon }{R}$.

The trajectory in the system $S$: The Lorentz transformations Eqs. 12.18 and 12.19, for the case of relative motion in the y-direction, read:

$\begin{array}{c}\left(i\right)\overline{x}=\gamma (x-\upsilon t)\\ \left(ii\right)\overline{y}=y\\ \left(iii\right)\overline{z}=z\\ \left(iv\right)\overline{t}=\gamma \left(t-\frac{\upsilon }{c^2}x\right)\end{array}$

Similarly,

$\begin{array}{c}(i\text{'})x=\gamma (\overline{x}-\upsilon \overline{t})\\ (ii\text{'})y=\overline{y}\\ (iii\text{'})z=\overline{z}\\ (iv\text{'})t=\gamma \left(\overline{t}-\frac{\upsilon }{c^2}\overline{x}\right)\end{array}$

$\begin{array}{c}\overline{x}=x,x=\overline{x}\\ \overline{y}=\gamma (y-\upsilon t),y=\gamma (\overline{y}-\upsilon \overline{t})\\ \overline{z}=z,z=\overline{z}\\ \overline{t}=\gamma \left(t-\frac{\upsilon }{c^2}y\right),t=\gamma \left(\overline{t}-\frac{\upsilon }{c^2}\overline{y}\right)\end{array}$

So, the trajectory in S is given by:

$\begin{array}{l}x=0\\ y=\gamma (-R\sin \omega \overline{t}+\upsilon \overline{t})\end{array}$

Putting the value of $\overline{t}$ and $t$,

$\begin{array}{c}y=\gamma \left\{-R\sin \left[[\omega \gamma \left(t-\frac{\upsilon }{c^2}y\right)\right]+\upsilon \gamma \left(t-\frac{\upsilon }{c^2}y\right)\right\}\\ y=-\gamma R\sin \left[\omega \gamma \left(t-\frac{\upsilon }{c^2}y\right)\right]+\upsilon \gamma t-\frac{{\upsilon }^2}{c^2}y\gamma \\ \gamma (y-\upsilon t)=-R\sin \left[\omega \gamma \left(t-\frac{\upsilon }{c^2}y\right)\right]\end{array}$

And,

$\begin{array}{l}z=R(1-{\cos }^2\omega \overline{t})\\ z=R\left[1-\cos \omega \gamma \left(t-\frac{\upsilon }{c^2}y\right)\right]\end{array}$

So,

$\begin{array}{c}x=0\\ y=\upsilon t-\frac{R}{\gamma }\sin \left[\omega \gamma \left(t-\frac{\upsilon }{c^2}y\right)\right]\\ z=R-R\cos \left[\omega \gamma \left[t-\frac{\upsilon }{c^2}\right]\right]\\ z-R=-R\cos \left[\omega \gamma \left(t-\frac{\upsilon }{c^2}\right)\right]\end{array}$

Eliminating the trigonometric terms by squaring and adding both equations,

$\begin{array}{c}{\gamma }^2{(y-\upsilon t)}^2+{(z-R)}^2=R^2{\sin }^2\left\{\left\{\omega \gamma \left(t-\frac{\upsilon }{c^2}y\right)\right\}+R^2{\cos }^2\left\{\{\omega \gamma \left(t-\frac{\upsilon }{c^2}y\right)\right\}\right\}\\ {\gamma }^2{(y-\upsilon t)}^2+{(z-R)}^2=R^2\left[{\sin }^2\left\{\omega \gamma \left(t-\frac{\upsilon }{c^2}y\right)\right\}+{\cos }^2\left\{\omega \gamma \left(t-\frac{\upsilon }{c^2}y\right)\right\}\right]\\ {\gamma }^2{(y-\upsilon t)}^2+{(z-R)}^2=R^2\end{array}$

If the above equation represents an elliptical cycloid with the horizontal axis stretched out.

The trajectory of the charged particle at different time intervals is represented below,

Hence, the trajectory of the path of the charge is ‘an elliptical cycloid’ represented by the equation ${\gamma }^2{(y-\upsilon t)}^2+{(z-R)}^2=R^2$.